Question: What is the density of 2 unknown liquids, and how precise are the measurements used to make the calculations? Introduction: Background Information: The density of a liquid, ? , is calculated by dividing mass by volume, m/v. This quantity, thus, is a measurement of the mass of an object per cubic unit. Precision can be defined as the accuracy of a measurement when compared to other measurements, not necessarily accurate when compared to the theoretical measurement.
In other words, all measurements made are within error range of each other, and are thus qualified as being precise. Materials: 2 unknown liquids: one blue and one green. A liquid of known density: ethyl alcohol. A 100 mL graduated cylinder A balance 2 250 mL beakers Method: 1. Put each unknown liquid into a beaker. 2. Bring the beakers to your workstation. 3. Put the empty graduated cylinder onto the balance, and tare it. 4. Pour 10 mL of one unknown liquid into the graduated cylinder. 5. Weigh the graduated cylinder. 6.
Repeat steps 4 and 5 until 50 mL have been poured and weighed. 7. Pour out the graduated cylinder, and wash and dry it. 8. Repeat steps 3-6 for the other unknown liquid and the ethyl alcohol. Data: Volume (mL) (note: this is only the objective volume) Liquid A (g) Liquid B (g) Isopropol Alcohol (g) Isopropol Alcohol: 16. 0 mL: 14. 032 g 27. 0 mL: 23. 679 g 44. 0 mL: 38. 588 g Discussion/ Observations: On one of the experimental trials, the data was slightly skewed. This occurred during the Liquid B measurements, during the 10 mL volume.
This likely affected the data to a slight degree for that trial. It should also be noted that the volume was not always exactly as shown in the chart. In cases where it was slightly different, I noted that in my graph for simplicity purposes. My graph does not have error lines because they were of an insignificant height, and would not have been useable to a reader. Conclusion: Volume was the limiting factor in accuracy during the experiment.
This was due to the fact that while the scale was able to determine the mass of the liquid to within . 01 g, i??.01, the volume was limited by the fact that it was more approximately determined, and thus had an error more closely representing i??. 3. Interpolation is the process of estimating what values are that lie within the data range for a specific set of numbers. This would utilize the slope of the data, and would then estimate what a y-value would be for a specific x, or vice-versa. Extrapolation is the opposite effect, and occurs when the slope is utilized to extend the data set, estimating what data values would be if more measurements had been taken such that the range was extended.
This would be a viable process for the above graph, as the numbers will only increase: there is no chance of decrease due to attributes such as mass or volume. The water measurements can be used as an indicator of accuracy and precision because water’s density is 1 mg/ mL, and thus you can determine how close your measurements are to the known value. They can also be used to determine precision, as each water subset should have the same density, so if the densities you measure are close to each other, then you are fairly precise. Based on their measurements, my groups was fairly precise in our measurement.